Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
const int MAX_N = 1e2 + 10;
int N, M, Nr[MAX_N][MAX_N];
int main() {
  cin >> N >> M;
  for (int(i) = 0; (i) < (int)(N); (i)++)
    for (int(j) = 0; (j) < (int)(M); (j)++) scanf("%d", &Nr[i][j]);
  vector<int> AnsN;
  vector<int> AnsM;
  if (N < M) {
    for (int(i) = 0; (i) < (int)(N); (i)++) {
      int minV = INF;
      for (int(j) = 0; (j) < (int)(M); (j)++) minV = min(minV, Nr[i][j]);
      for (int(j) = 0; (j) < (int)(M); (j)++) Nr[i][j] -= minV;
      for (int(j) = 0; (j) < (int)(minV); (j)++) AnsN.push_back(i);
    }
    for (int(i) = 0; (i) < (int)(M); (i)++) {
      int minV = INF;
      for (int(j) = 0; (j) < (int)(N); (j)++) minV = min(minV, Nr[j][i]);
      for (int(j) = 0; (j) < (int)(N); (j)++) Nr[j][i] -= minV;
      for (int(j) = 0; (j) < (int)(minV); (j)++) AnsM.push_back(i);
    }
  } else {
    for (int(i) = 0; (i) < (int)(M); (i)++) {
      int minV = INF;
      for (int(j) = 0; (j) < (int)(N); (j)++) minV = min(minV, Nr[j][i]);
      for (int(j) = 0; (j) < (int)(N); (j)++) Nr[j][i] -= minV;
      for (int(j) = 0; (j) < (int)(minV); (j)++) AnsM.push_back(i);
    }
    for (int(i) = 0; (i) < (int)(N); (i)++) {
      int minV = INF;
      for (int(j) = 0; (j) < (int)(M); (j)++) minV = min(minV, Nr[i][j]);
      for (int(j) = 0; (j) < (int)(M); (j)++) Nr[i][j] -= minV;
      for (int(j) = 0; (j) < (int)(minV); (j)++) AnsN.push_back(i);
    }
  }
  bool chk = true;
  for (int(i) = 0; (i) < (int)(N); (i)++)
    for (int(j) = 0; (j) < (int)(M); (j)++)
      if (Nr[i][j] != 0) chk = false;
  if (!chk) {
    puts("-1");
    return 0;
  }
  printf("%d\n", ((int)(AnsN).size()) + ((int)(AnsM).size()));
  for (int x : AnsN) printf("row %d\n", x + 1);
  for (int x : AnsM) printf("col %d\n", x + 1);
  return 0;
}